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chemistry, please help

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2NO2(g)--> 2NO(g) + O2(g) , H=+114.2kJ

(Note: H, S, G all have a degree sign next to them)

NO: H(enthalpy)=90.3kJ/mol, S(entropy)=210.7J/mol*K, G(gibbs energy)=86.6

O2: H(enthalpy)=0, S(entropy)=?, G(gibbs energy)=0 kJ/mol

NO2: H(enthalpy)=33.2, S=239.9, G=51

For part a, I have to calcuate G for the reaction.

I used the Gibbs free standard reaction formula:where i have to find the sum of products minus the sum of reactants.
I got 71.2 kJ/mol. I know I got that right because my friend got the same answer too.

for part b, I have to calculate standard entropy, S for only O2.
There's a table in my book saying that entropy of O2 is 205 J/mol*K.
But I have to solve this using formulas and I was wondering if you can help on how to approach this problem.

  • i forgot to write down another part of question -

    The temperature of the reaction is 298 kelvin

  • tried solving it, please check answer -

    This is what I did:

    my teacher said that this is a two step calculation:
    First step:equation of heat
    m=mass of O2=32 grams
    T=298+273=573 Celsius
    q=32 g of O2(4.184J/g C)(573 C)

    then I used the equation:
    S=76450J/mol / (298 K)
    =257 J/mol*K

    I looked at a table in my chemistry book, and it says that S of O2 is 205 J/mol*K.
    I'm assuming that's the answer I should reach, but I didn't.

    So please see what I did wrong.
    Thank you

  • chemistry, please help -

    298 K is not 573 C. It is 25 C

    Entropy at 298 requires an integration of C dT/T from 0 K to 298, including solid and liquid phases. It cannot be calculated with a simple formula. The 205 J/mol K number for 298 K agrees with my JANNAF Tables.

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